number of default class A

the default cl ass A mask is 255.0.0.0 that leaves 8 bits for network. so to calculate number of networks in class A network we raise 2 to th e power of 8, class B 2 to the power of 14 and class C 2 to the power of 21.

why not 2 to the power of 8 for class A and 2 to the power of 16 for class B and 2 to the power of 24 for class C

Class A has the classfull mask of 255.0.0.0 yes, But there is also another restriction. A class A address must start with the 0 bit at the first octet, so the bits must look like this:

0YYYYYYY.XXXXXXXX.XXXXXXXX.XXXXXXXX

Were Y are network bits and X are host bits. This give class A the following range:

00000000.XXXXXXXX.XXXXXXXX.XXXXXXXX to 01111111.XXXXXXXX.XXXXXXXX.XXXXXXXX

In Decimals: 0.X.X.X to 127.X.X.X

As you can see, it is 2 to the power 7 (8-1) = 128 networks, since the 8th bit is fixed to 0. These are 128 networks for class A. And you have to subtract network zero and the private/reserved ranges.

For class B, The first bit must be 1 and the second bit must be 0. Like this:

10YYYYYY.YYYYYYYY.XXXXXXXX.XXXXXXXX

So the range (with a mask of 255.255.0.0) will be:

10000000.00000000.XXXXXXXX.XXXXXXXX to 10111111.11111111.XXXXXXXX.XXXXXXXX

In Decimals: 128.0.X.X to 191.255.X.X

This gives Class B 2 to the power 14 (16-2) networks. Again you have exclude private ranges.

For class C, the bits should be:

110YYYYY.YYYYYYYY.YYYYYYYY.XXXXXXXX

So:

11000000.00000000.00000000.XXXXXXXX to 11011111.11111111.11111111.XXXXXXXX

In Decimal: 192.0.0.X to 223.255.255.X

A total of 2 to the power 21 (24-3).

I can give you the details after subtracting private ranges, but the above is enough for your question I guess . Hope it helps.